Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (a)

MATH105 April 2015

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Question 05 (a)
Suppose that ${\frac {df}{dx}}={\frac {x}{1+3x^{3}}}$ and $f(0)=1.$ Find the Maclaurin series for $f(x).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Expand ${\frac {df}{dx}}$ as a Maclaurin series by using the geometric series ${\frac {1}{1+y}}={\frac {1}{1-(-y)}}=1-y+y^{2}-y^{3}+\cdots =\sum _{n=0}^{\infty }(-y)^{n}.$

Hint 2
Once a series for ${\frac {df}{dx}}$ is obtained, integrate it termwise and use the condition $f(0)=1$ to obtain a series for $f(x)$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since ${\frac {df}{dx}}=x\left({\frac {1}{1+3x^{3}}}\right)$, a Maclaurin series for ${\frac {df}{dx}}$ can be computed from the geometric series ${\frac {1}{1+y}}={\frac {1}{1-(-y)}}=\sum _{n=0}^{\infty }(-y)^{n}=\sum _{n=0}^{\infty }(-1)^{n}y^{n}.$ By substituting $y=3x^{3}$, we obtain: ${\frac {df}{dx}}=x\left({\frac {1}{1+3x^{3}}}\right)=x\sum _{n=0}^{\infty }(-1)^{n}(3x^{3})^{n}=\sum _{n=0}^{\infty }(-3)^{n}x^{3n+1}.$ To obtain a series for $f(x)$, we integrate the series for ${\frac {df}{dx}}$ term by term and apply the initial condition $f(0)=1$. Integrating the series for ${\frac {df}{dx}}$ implies that $f(x)=\sum _{n=0}^{\infty }(-3)^{n}\int x^{3n+1}\,dx=\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}+C$ for some constant C determined by the initial condition. Since $\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}$ evaluates to 0 at $x=0$, we must have $C=1$ for $f(0)=1$. Therefore, a Maclaurin series for $f(x)$ is given by ${\color {blue}1+\sum _{n=0}^{\infty }{\frac {(-3)^{n}x^{3n+2}}{3n+2}}=1+{\frac {x^{2}}{2}}-{\frac {3x^{5}}{5}}+{\frac {9x^{8}}{8}}+\cdots .}$